anonymous

4 years ago

Is space curvature relative to the frame of reference? If I have an object that is of some length \(l\) moving at a relativistic velocity \(v\) for some reference frame, then length contraction states that \(l=\gamma \times l'\). But at the frame of reference, there is no length contraction. The curvature of space should be Euclidean. However, relative to the moving object in the prior example, the exact opposite is true, with there being no curvature near the primed frame from the primed point of view. What if one of the objects is accelerating? If I stand on a planet, will my geometry of space agree with someone who's not standing on one?